Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix E - Sigma Notation - E Exercises: 31

Answer

The solution is $$\sum_{i=1}^n(i^2+3i+4)=\frac{1}{3}n\left(n^2+6n+17\right).$$

Work Step by Step

First, we will use the fact that the sum passes through terms individually, as well as the fact that the constant can be put in front of the sum: $$A=\sum_{i=1}^n(i^2+3i+4)=\sum_{i=1}^ni^2+3\sum_{i=1}^ni+\sum_{i=1}^n4.$$ Now, using equalities from Theorem 3 we get $$A=\frac{n(n+1)(2n+ 1)}{6}+3\frac{n(n+1)}{2}+4n=\\ n\left(\frac{(n+1)(2n+1)}{6}+\frac{3(n+1)}{2}+4\right)=n\left((n+1)\left(\frac{2n+1}{6}+\frac{3}{2}\right)+4\right)=n\left((n+1)\left(\frac{2n+1}{6}+\frac{3}{2}\right)+4\right)=n\left((n+1)\frac{2n+1+9}{6}+4\right)=n\left(\frac{(n+1)(n+5)}{3}+4\right)=\frac{1}{3}n\left(n^2+6n+17\right).$$
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